Integer equals Ceiling iff between Number and One More

Theorem

Let $x \in \R$ be a real number.

Let $\ceiling x$ be the ceiling of $x$.

Let $n \in \Z$ be an integer.

Then:

$\ceiling x = n \iff x \le n < x + 1$

Proof

Necessary Condition

Let $x \le n < x + 1$.

From $x \le n$, we have by Number not greater than Integer iff Ceiling not greater than Integer:

$\ceiling x \le n$

From $n < x + 1$:

$n - 1 < x$

Hence by Number greater than Integer iff Ceiling greater than Integer:

$n - 1 < \ceiling x$

We have that:

$\forall m, n \in \Z: m < n \iff m \le n - 1$

and so:

$n \le \ceiling x$

Thus as:

$\ceiling x \le n$

and:

$\ceiling x \ge n$

it follows that:

$\ceiling x = n$

$\Box$

Sufficient Condition

Let $\ceiling x = n$.

Then:

$\ceiling x \le n$

By Number not greater than Integer iff Ceiling not greater than Integer:

$x \le n$

From Number is between Ceiling and One Less:

$\ceiling x - 1 < x$

and so adding $1$ to both sides:

$\ceiling x < x + 1$

and so by hypothesis:

$n < x + 1$

So:

$\ceiling x = n \implies x \le n < x + 1$

$\Box$

Hence the result:

$\ceiling x = n \iff x \le n < x + 1$

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(f)}$